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Rigid Body Statics

Central Force Systems: Forces with a Common Point of Application

In engineering mechanics, the analysis of forces acting on an object plays a central role. An important special case is the central force system, where all these forces have a common point of application. In this chapter, we will explore the properties and analysis of central force systems and explore:

  • Definition and Properties: What is a central force system? What special properties distinguish it?
  • Basic Tasks of Rigid Body Statics for Central Force Systems:
    • Reduction: Reducing a system of multiple forces to a single point (center of force).
    • Equilibrium: Determining whether a body remains at rest or moves under the action of forces.
    • Decomposition: Decomposing a force into several individual forces.

Are you ready to unlock these secrets?

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Practice Exercises
Table of Contents

5. Second Basic Task: Equilibrium

Analytical Solution
Okay, aspiring engineer, pay attention!

The graphical method from the last chapter was pretty cool for 2D, but unfortunately it's not really accurate or practical for 3D. That's why we're switching gears now and heading into the mathematical fast lane: we'll solve equilibrium problems precisely and elegantly with pure calculation!

Grab your pen, paper, and your best buddy, the calculator, and let's go!

Preparation

Remember the insight from the last chapter? A central force system is in equilibrium when the sum of all forces is 0. Vectorially expressed:

$$ \begin{align} \tag{1} \vec{R} &=\sum\ \vec{F}_{i} = 0 \end{align} $$

But when is a vector actually 0?

What you need to know:
  • Forces are vectors. So in the Cartesian coordinate system, they have components for each axis:
    $$ \begin{align} \tag{2} \vec{F} &= \begin{pmatrix} F_x\\ F_y \\ F_z \end{pmatrix} \end{align} $$
  • Vectors are added component-wise. The x-component of the resultant, for example:
    $$ \begin{align} \tag{3} F_x &= {F_1}_x + {F_2}_x + \ldots + {F_n}_x = \sum_{i=1}^n {F_i}_x \end{align} $$
  • A vector is 0 if all its components are 0..
    $$ \begin{align} \tag{4} \vec{R} &=\sum\ \vec{F}_{i} = 0 = \begin{pmatrix} 0\\ 0 \\ 0 \end{pmatrix} \end{align} $$
The "Scalar Equilibrium Conditions"

This results in the following formulas, which must be fulfilled as conditions for equilibrium in a central force system in the Cartesian coordinate system:

  • 2D:
    $$ \begin{aligned} R_x &= \sum_{i=1}^n F_{i_x} = 0 \\[12pt] R_y &= \sum_{i=1}^n F_{i_y} = 0 \end{aligned} $$
  • 3D:
    $$ \begin{aligned} R_x &= \sum_{i=1}^n F_{i_x} = 0 \\[12pt] R_y &= \sum_{i=1}^n F_{i_y} = 0 \\[12pt] R_z &= \sum_{i=1}^n F_{i_z} = 0 \end{aligned} $$

(2.7)

Meaning: To determine the equilibrium of a central force system, we need two equations (2D) or three equations (3D)..

How it works step by step:

  1. Break down the forces: Each force Fi can be broken down into two (three) components: a horizontal Fix and a vertical Fiy (and in 3D also Fiz).

    Tip: To determine the components, do not use the direction angle \(\alpha_i\), but rather one of the acute angles of the force triangle between the force vector and its components, e.g. \(\alpha_{i_{acute}}\).

    This Figure 3.5.1 shows a force vector F in the 4 different quadrants of the plane coordinate system.
    Fig. 3.5.1: Four Cases: Direction angle \(\alpha_i\) and acute angle \(\alpha_{i_{acute}}\)
    Why?
    • Easy calculation: Calculate the angle, done! No stress with reference directions and complicated formulas.
    • Super visual: The acute angle is unambiguous and can be easily represented - perfect for grasping the relationships in the system at a glance.

    But be careful: When setting up equations (2.7), pay attention to the positive direction! Forces in this direction get a positive sign, forces in the opposite direction get a negative sign.

    Extra Tip: Since the positive directions are freely selectable, they are often marked with arrows. So instead of "\(\sum F_{i_x}=0\) positive to the right" you simply write "\(\rightarrow:\)".

  2. Add the components: Sum all horizontal and vertical components separately: \(\sum F_{i_x}= \dots\) and \(\sum F_{i_y}= \dots\)
  3. Evaluation:
    • Equilibrium reached: If all equations \((\sum F_{i_x}=0\) and \(\sum F_{i_y}=0\)) are satisfied, the system is in equilibrium.
    • No equilibrium: If at least one equation does not meet this condition, there is no equilibrium.
See, it's that simple!

With this method, you can crack any central force system. Try it out now and become a master of central force systems!

But beware: This method only applies to central force systems (lines of action through one point). With other force arrangements, you may need more sophisticated methods.

However, with a little practice and enjoyment of equilibrium calculations, you'll quickly become a pro!

Any questions?

No problem! In the next chapters, we'll delve deeper into the world of central force systems and unlock further secrets.

Stay tuned and good luck!